Let d(n) represent the number of diagonals of a polygon of n sides. Here is a table of values of d(n).
n 3 4 5 6 7 8 9 10 11
d(n)0 2 5 9 14 20 27 35 44
Find a polynomial formula for d(n) in terms of n
check the differences:
1st: 2 3 4 5 6 ...
2nd: 1 1 1 1 1
So, it will be a quadratic
d = an^2+bn+c
Plug in three points to find a,b,c
To find a polynomial formula for d(n) in terms of n, we need to find a pattern in the given values. Let's examine the differences between consecutive terms of d(n):
d(n): 0 2 5 9 14 20 27 35 44
Difference: 2 3 4 5 6 7 8 9
We observe that the differences between consecutive terms are increasing by 1 each time. This suggests that d(n) might be a polynomial of degree 2.
To find the polynomial, we can start by assuming that d(n) is of the form:
d(n) = an^2 + bn + c
To solve for the coefficients a, b, and c, we can substitute the values of d(n) and n from the table into the equation. We get the following system of equations:
1 = a(3^2) + b(3) + c
2 = a(4^2) + b(4) + c
5 = a(5^2) + b(5) + c
Simplifying these equations, we have:
9a + 3b + c = 1
16a + 4b + c = 2
25a + 5b + c = 5
To solve this system of equations, we can use various methods such as substitution, elimination, or matrix methods. Let's use elimination:
Subtracting the first equation from the second equation, we get:
7a + b = 1
Subtracting the second equation from the third equation, we get:
9a + b = 3
Solving these two equations gives us:
a = 1/2
b = -1/2
Substituting these values back into the first equation, we get:
9(1/2) - 3(1/2) + c = 1
4.5 - 1.5 + c = 1
c = -4
Therefore, the polynomial formula for d(n) in terms of n is:
d(n) = (1/2)n^2 - (1/2)n - 4